# 3.1 Simple Interest: Principal, Rate, Time

## LEARNING OBJECTIVES

- Understand the concept of simple interest.
- Calculate interest, principal, rate, and time for simple interest transactions.

**Interest** is the fee paid by a borrower to a lender for using the lender’s money. As a borrower, the interest paid is an expense. But as a lender or an investor, interest earned is income. There are two types of interest: simple interest and compound interest.

In a simple interest environment, interest is calculated solely on the initial amount of money invested or borrowed at the beginning of the transaction at a specified simple interest rate for the entire time period. A loan or investment always involves two parties—one giving and one receiving. No matter which party you are in the transaction, the amount of interest remains unchanged. The only difference lies in whether you are earning or paying the interest.

Suppose $1,000 is placed into an account with 12% simple interest for a period of 12 months. For the entire term of this transaction, the amount of money in the account always equals $1,000. During this period, interest accrues at a rate of 12%, but the interest is never placed into the account. When the transaction ends after 12 months, the $120 of interest and the initial $1,000 are then combined to total $1,120.

## The Simple Interest Formula

The amount of simple interest ([latex]I[/latex]) is calculated as a percent of the amount of money invested or borrow over a specified period of time.

[latex]\begin{eqnarray*} I & = & P \times r \times t \end{eqnarray*}[/latex]

where

**[latex]I[/latex] is the amount of interest.**This is the dollar amount of interest paid or received.**[latex]P[/latex] is the present value or principal.**This is the amount borrowed or invested at the beginning of the time period.**[latex]r[/latex] is the simple interest rate.**This is the rate of interest that is charged or earned during a specified time period. The simple interest rate is expressed as a percent for a given time period, usually annually or per year, unless otherwise specified.**[latex]t[/latex] is the period of time or term of the investment or loan.**The time period is the length of time for which interest is earned or charged on the investment or loan respectively.

### NOTE

Recall that algebraic equations require all terms to be expressed with a common unit. This principle remains true for the simple interest formula, particularly with regard to the interest rate and the time period. For example, if you have a 3% annual simple interest rate for nine months, then either

- The time needs to be expressed annually as [latex]\frac{9}{12}[/latex] of a year to match the yearly interest rate, or
- The interest rate needs to be expressed monthly as [latex]\frac{3\%}{12}=0.25\%[/latex] per month to match the number of months.

It does not matter which conversion you do as long as you express both the interest rate and the time in the same unit. If one of these two variables is your algebraic unknown, the unit of the known variable determines the unit of the unknown variable. For example, assume that you are solving the simple interest formula for the time period. If the interest rate used in the formula is annual, then the time period is expressed in number of years.

## EXAMPLE

Julio borrowed $1,100 from Maria five months ago. When he first borrowed the money, they agreed that he would pay Maria 5% simple interest. If Julio pays her back today, how much interest does he owe her?

**Solution:**

**Step 1:** The given information is

[latex]\begin{eqnarray*} P & = & \$1,100 \\ r & = & 5\% \mbox{ (per year)} \\ & = & 0.05 \\ t & = & 5 \mbox{ months}\end{eqnarray*}[/latex]

**Step 2:** The interest rate is annual, but the time is in months. Convert the time period from months to years.

[latex]\displaystyle{t=\frac{5}{12}}[/latex]

**Step 3:** Solve for the amount of interest, [latex]I[/latex].

[latex]\begin{eqnarray*} I & = & P \times r \times t \\ & = & 1,100 \times 0.05 \times \frac{5}{12} \\ & = & \$22.92 \end{eqnarray*}[/latex]

For Julio to pay back Maria, he must reimburse her for the $1,100 principal borrowed plus an additional $22.92 of simple interest as per their agreement.

## Solving for [latex]P[/latex], [latex]r[/latex], or [latex]t[/latex]

Four variables are involved in the simple interest formula [latex]I =P \times r \times t[/latex], which means that any three can be known quantities and require you to solve for the fourth missing variable.

## EXAMPLE

What amount of money invested at 6% annual simple interest for 11 months earns $2,035 of interest?

**Solution:**

**Step 1:** The given information is

[latex]\begin{eqnarray*} I & = & \$2,035 \\ r & = & 6\% \mbox{ (per year)} \\ & = & 0.06 \\ t & = & 11 \mbox{ months}\end{eqnarray*}[/latex]

**Step 2:** The interest rate is annual, but the time is in months. Convert the time period from months to years.

[latex]\displaystyle{t=\frac{11}{12}}[/latex]

**Step 3:** Solve for the principal [latex]P[/latex].

[latex]\begin{eqnarray*} P & = & \frac{I}{ r \times t }\\ & = & \frac{2,035} { 0.06 \times \frac{11}{12}} \\ & = & \$37,000 \end{eqnarray*}[/latex]

To generate $2,035 of simple interest at 6% over a time frame of 11 months, $37,000 must be invested.

## EXAMPLE

For how many months must $95,000 be invested to earn $1,187.50 of simple interest at an interest rate of 5%?

**Solution:**

**Step 1:** The given information is

[latex]\begin{eqnarray*} P & = & \$95,000 \\ I & = & \$1,187.50 \\ r & = & 5\% \mbox{ (per year)} \\ & = & 0.05 \end{eqnarray*}[/latex]

**Step 2:** Solve for the time period [latex]t[/latex].

[latex]\begin{eqnarray*} t & = & \frac{I}{P \times r }\\ & = & \frac{1,187.50} { 95,000 \times 0.05} \\ & = & 0.25 \end{eqnarray*}[/latex]

**Step 3:** The time period found in step 2 is in years, so convert the years to months by multiplying by 12.

[latex]\begin{eqnarray*} \mbox{Time period in months} & = & 0.25 \times 12 \\ & = & 3 \end{eqnarray*}[/latex]

For $95,000 to earn $1,187.50 at 5% simple interest, it must be invested for a three-month period.

## TRY IT

If you want to earn $1,000 of simple interest at a rate of 7% in a span of five months, how much money must you invest?

**Click to see Solution**

[latex]\begin{eqnarray*} P & = & \frac{I}{r \times t} \\ & = & \frac{1000}{0.07 \times \frac{5}{12}} \\ & = & \$34,285.71 \end{eqnarray*}[/latex]

## TRY IT

If you placed $2,000 into an investment account earning 3% simple interest, how many months does it take for you to have $2,025 in your account?

**Click to see Solution**

[latex]\begin{eqnarray*} t & = & \frac{I}{ P \times r} \\ & = & \frac{25}{2000 \times 0.03} \\ & = & 0.4166... \\ \\ \mbox{Time in months} & = & 0.04166... \times 12 \\ & = & 5 \end{eqnarray*}[/latex]

## TRY IT

A $3,500 investment earned $70 of interest over the course of six months. What annual rate of simple interest did the investment earn?

**Click to see Solution**

[latex]\begin{eqnarray*} r & = & \frac{I}{ P \times t} \\ & = & \frac{70}{3500 \times \frac{6}{12}} \\ & = & 0.04 \\ & = & 4\% \end{eqnarray*}[/latex]

## Time and Dates

In the examples of simple interest so far, the time period was given in months. While this is convenient in many situations, financial institutions and organizations calculate interest based on the exact number of days in the transaction, which changes the interest amount.

To illustrate this, assume you had money saved for the entire months of July and August, where [latex]t = \frac{2}{12}[/latex] or [latex]t = 0.16666...=0.1\overline{6}[/latex] of a year. However, if you use the exact number of days, the 31 days in July and 31 days in August total 62 days. In a 365-day year that is [latex]t =\frac{62}{365}[/latex] or t = 0.169863 of a year. Notice a difference of 0.003196 (0.169863 – 0.16) occurs. Therefore, to be precise in performing simple interest calculations, you must calculate the **exact number of days** involved in the transaction. Although you can count the exact number of days by hand, most financial calculators and spreadsheet software like Excel have built-in functions to calculate the number of days between two dates.

## USING THE TI BAII PLUS CALCULATOR TO COUNT THE NUMBER OF DAYS BETWEEN DATES.

To count the number of days between two dates:

- Press 2nd DATE (the 1 button) to enter the date worksheet.
- At the DT1 screen, enter the first date. Dates are entered in the form mm.ddyy. For example, enter May 19, 2023 as 05.1923. After entering the date press ENTER.
- Press the down arrow.
- At the DT2 screen, enter the second date. Dates are entered in the form mm.ddyy. For example, enter August 7, 2023 as 08.0723. After entering the date press ENTER.
- Press the down arrow.
- At the DBD screen press CPT to calculate the number of days between the two entered dates. For example, there are 80 days between May 19, 2023 and August 7, 2023.

_{Calculating Dates and Days (Month-Day-Year) by Joshua Emmanuel [3:32] (transcript available).}

### NOTE

When solving for [latex]t[/latex], decimals may appear in your solution. For example, if calculating [latex]t[/latex] in days, the answer may show up as 45.9978 or 46.0023 days. However, interest is calculated only on the complete number of days. This occurs because the interest amount, [latex]I[/latex], used in the calculation has been rounded off to two decimals. Because the interest amount is imprecise, the calculation of [latex]t[/latex] is imprecise. When this occurs, round [latex]t[/latex] off to the nearest integer.

## EXAMPLE

Mark borrowed $4,200 from his friend Chris on November 3, 2022. Their agreement required that Mark pay back the loan on April 24, 2023 with 8% simple interest. How much interest did Mark pay Chris?

**Solution:**

**Step 1:** The given information is

[latex]\begin{eqnarray*} P & = & \$4,200 \\ r & = & 8\% \mbox{ (per year)} \\ & = & 0.08 \\ t & = & \frac{294}{365} \end{eqnarray*}[/latex]

**Step 2:** Solve for the intererst.

[latex]\begin{eqnarray*} I & = & P \times r \times t \\ & = & 4200 \times 0.08 \times \frac{294}{365} \\ & = & $270.64 \end{eqnarray*}[/latex]

Mark pays Chris $270.64 in interest.

## EXAMPLE

On September 13, 2011, Aladdin decided to pay back the Genie on his loan of $15,000 at 9% simple interest. If he paid the Genie the principal plus $1,283.42 of interest, on what day did he borrow the money from the Genie?

**Solution:**

**Step 1:** The given information is

[latex]\begin{eqnarray*} P & = & \$15,000 \\ I & = & \$1,283.42 \\ r & = & 9\% \mbox{ (per year)} \\ & = & 0.09 \end{eqnarray*}[/latex]

**Step 2:** Solve for the time period.

[latex]\begin{eqnarray*} t & = & \frac{I}{P \times r }\\ & = & \frac{1,283.42} { 15,000 \times 0.09} \\ & = & 0.95068... \end{eqnarray*}[/latex]

**Step 3:** The time period found in step 2 is in years, so convert the years to days by multiplying by 365.

[latex]\begin{eqnarray*} \mbox{Time period in days} & = & 0.95068... \times 365 \\ & = & 346.998... \\ & \rightarrow& 347 \end{eqnarray*}[/latex]

**Step 4:** Use the DATE function to calculate the start date (DT1).

- Press 2nd DATE.
- Press the down arrow to skip the DT1 screen and move to the DT2 screen.
- At the DT2 screen enter 09.1311 for September 13, 2011. Press ENTER.
- Press the down arrow.
- At the DBD screen, enter 347 and press ENTER.
- Press the up arrow twice to return the the DT1 screen.
- At the DT1 screen, press CPT. The date is October 1, 2010.

If Aladdin owed the Genie $1,283.42 of simple interest at 9% on a principal of $15,000, he must have borrowed the money 347 days earlier, which is October 1, 2010.

## TRY IT

Brynn borrowed $25,000 at 1% per month from a family friend to start her entrepreneurial venture on December 2, 2011. If she paid back the loan on June 16, 2012, how much simple interest did she pay?

**Click to see Solution**

[latex]\begin{eqnarray*} I & = & P \times r \times t \\ & = & 25,000 \times 0.12 \times \frac{197}{365} \\ & = & \$1,619.18 \end{eqnarray*}[/latex]

## TRY IT

If $6,000 principal plus $132.90 of simple interest was withdrawn on August 14, 2011, from an investment earning 5.5% interest, on what day was the money invested?

**Click to see Solution**

[latex]\begin{eqnarray*} t & = & \frac{I}{P \times r} \\ & = & \frac{132.90}{6000 \times 0.055} \\ & = & 0.40272... \\ \\ \mbox{Time in days} & = & 0.40272... \times 365 \\ & = & 146.9954... \\ & \rightarrow & 147 \end{eqnarray*}[/latex]

The money was invested on March 20, 2011.

**Exercises**

- Brynn borrowed $25,000 at 1% per month from a family friend to start her entrepreneurial venture on December 2, 2011. If she paid back the loan on June 16, 2012, how much simple interest did she pay?

**Click to see Answer**$1,619.18

- How much simple interest is earned on $50,000 over 320 days if the interest rate is:
- 3%
- 6%
- 9%

**Click to see Answer**a. $1,315.07, b. $2,630.14, c. $3,945.21

- If you placed $2,000 into an investment account earning 3% simple interest, how many months does it take for you to have $2,025 in your account?

**Click to see Answer**5 months

- If you want to earn $1,000 of simple interest at a rate of 7% in a span of five months, how much money must you invest?

**Click to see Answer**$34,285.71

- If $6,000 principal plus $132.90 of simple interest was withdrawn on August 14, 2011, from an investment earning 5.5% interest, on what day was the money invested?

**Click to see Answer**March 20, 2011

- Jessica decided to invest her $11,000 in two back-to-back three-month term deposits. On the first three-month term, she earned $110 of interest. If she placed both the principal and the interest into the second three-month term deposit and earned $145.82 of interest, how much higher or lower was the interest rate on the second term deposit?

**Click to see Answer**1.25% higher

- Marrina is searching for the best way to invest her $10,000. One financial institution offers 4.25% on three-month term deposits and 4.5% on six-month term deposits. Marrina is considering either doing two back-to-back three-month term deposits or just taking the six-month deposit. It is almost certain that interest rates will rise by 0.5% before her first threemonth term is up. She will place the simple interest and principal from the first three-month term deposit into the second three-month deposit. Which option should Marrina pursue? How much better is your recommended option?

**Click to see Answer**Back-to-back 3 month investments, $1.26 more

- Evaluate each of the following $10,000 investment alternatives and recommend the best alternative for investing any principal regardless of the actual amount. Assume in all cases that the principal and simple interest earned in prior terms are placed into subsequent investments.
**Alternative 1:**6% for 1 year**Alternative 2:**5% for 6 months, then 7% for 6 months**Alternative 3:**5.25% for 3 months, then 5.75% for 3 months, then 6.25% for 3 months, then 6.75% for 3 months

What percentage more interest is earned in the best alternative versus the worst alternative?

**Click to see Answer**Alternative 3, 0.1283% more than Alternative 1

#### Attribution

“8.1: Simple Interest: Principle, Rate, Time” from Business Math: A Step-by-Step Handbook Abridged by Sanja Krajisnik; Carol Leppinen; and Jelena Loncar-Vines is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

“8.1: Principal, Rate, Time” from Business Math: A Step-by-Step Handbook (2021B) by J. Olivier and Lyryx Learning Inc. through a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License unless otherwise noted.